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kakasveta [241]
3 years ago
5

Who was the first person who went to Mars?

Chemistry
2 answers:
Novay_Z [31]3 years ago
8 0

Answer:Wernher von Braun proposal (1947 through 1950s) Wernher von Braun was the first person to make a detailed technical study of a Mars mission.

Explanation:

vekshin13 years ago
7 0
Eli Cologne was the first person who went to mars
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Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen- containing comp
dimulka [17.4K]

Answer:

1.1 × 10² g

Explanation:

First, we will convert 1.0 L to cubic centimeters.

1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³

The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:

1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g

1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:

1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g

4 0
3 years ago
Write a story explaining what happens to a drop of water during the water cycle
svlad2 [7]
Water was bored. He decided to go through the water cycle. He flew in the air as gas, then condensed into water again, and then rolled down into a river, only to find out that he would be stuck doing it forever
. The end
3 0
3 years ago
The table below shows the density of a sample of a mystery liquid you tested in the lab. Can you infer the identity of the subst
Mashutka [201]

Answer:

A. Yes, the substance must be water.

Explanation:

The density of a substance is unique to it. Density is defined the as the amount of substance contained per volume.

One of the ways of identifying a substance is to determine its density. Every matter is known to have their own specific densities. This makes them different from other substances. The density of gold is unique to it and it differs from that of silver.

In fact, water has density of 1.00gcm⁻³. Experimental errors and some little factors must have altered our expected figure. This a case of precision and accuracy in the experiment.

7 0
3 years ago
Why the measured pressure of a gas under conditions that are very close to those that would result in condensation will be lower
Snowcat [4.5K]

Answer:

Inter-molecular forces and molecular volumes are the chief reasons for lower measured pressure

Explanation:

The kinetic theory assumes that gas particles occupy a negligible fraction of the total volume of the gas. It also assumes that the force of attraction between gas molecules is zero.

However, during high pressure, the volume of the gas particles are not negligible compare to the total gas volume and as such the volume of a real gas under such condition is higher than the Ideal gas. Vander-waal attempted to modify the ideal gas equation by subtracting the excess volume from the ideal equation. The increased volume is the reason the measured pressure of a real gas is  less than an ideal gas

On the other hand,  close to condensation, the other  assumption of negligible forces of attraction becomes invalid. As inter-molecular distances decrease, inter-molecular forces increase reducing the bombardment of the wall of the container due to restricted particle movement and lower measured gas pressure.

3 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
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